If `c lt 0` and `ax^2 + bx + c = 0` does not have any real roots then prove that:
(i) `a-b + c lt 0`, (ii) `9a +3b +c lt 0`

Answer

To solve the problem, we need to prove two inequalities based on the given conditions. Let's break it down step by step.

### Given:
1. \( c < 0 \)
2. The quadratic equation \( ax^2 + bx + c = 0 \) does not have any real roots.

### Step 1: Understanding the condition for no real roots
For the quadratic equation \( ax^2 + bx + c = 0 \) to not have any real roots, the discriminant must be less than zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Since the equation does not have real roots, we have:
\[ b^2 - 4ac < 0 \]
This implies:
\[ b^2 < 4ac \]

### Step 2: Analyzing the implications of \( c < 0 \)
Since \( c < 0 \), we can rewrite the inequality \( b^2 < 4ac \) as:
\[ b^2 < 4a(c) \]
Given that \( c \) is negative, \( 4ac \) will also be negative if \( a > 0 \). However, since the parabola does not intersect the x-axis, we conclude that \( a < 0 \) (the parabola opens downwards).

### Step 3: Proving \( a - b + c < 0 \)
We will evaluate the quadratic function at \( x = -1 \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Since the quadratic function \( f(x) \) is negative for all \( x \) (as it has no real roots), we have:
\[ f(-1) < 0 \]
Thus:
\[ a - b + c < 0 \]

### Step 4: Proving \( 9a + 3b + c < 0 \)
Next, we evaluate the quadratic function at \( x = 3 \):
\[ f(3) = a(3)^2 + b(3) + c = 9a + 3b + c \]
Again, since the quadratic function \( f(x) \) is negative for all \( x \), we have:
\[ f(3) < 0 \]
Thus:
\[ 9a + 3b + c < 0 \]

### Conclusion
We have proven both inequalities:
1. \( a - b + c < 0 \)
2. \( 9a + 3b + c < 0 \)